Differential Index

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The differential index of a DAE is the number of times $\frac{d}{dt}$ must be applied in order to turn the DAE into an ODE.

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General Method

1. Check if $\partial c/\partial z$ is full rank
2. If not, apply $d/dt$ to the
3. Check if …
4. … See lecture 11

Hard VS Easy DAE’s

Easy DAE: Differential index = $1$ Hard DAE: Differential index > $1$

• Aka if $0=c(z,x)$ cannot be used to find $z$.

Note

Constrained Lagrange yields only hard DAE’s.

• This is why we had to differentiate the constraint of the pendulum example (i.e to reduce its index to $1$).
  • [[New Lagrangian#pendulum-without—theta|Pendulum without $theta$]]

Examples

A General Example

Given the following relation

$$\begin{array}{rl}{\dot{x}}_1& =x_2\\ {\dot{x}}_2& =x_3\\ & ⋮\\ {\dot{x}}_n& =f(x,z)\\ 0& =c(x_1)\end{array}$$

Since we cannot use $c(x_1)=0$ to find $z$, then $\frac{\partial c}{\partial z}(=0\text{ in the scalar case})$ is not full rank.

We then find $\frac{d}{dt}c(x_1)=\dot{c}(x_1,x_2)$, still not full rank ($\frac{\partial \dot{c}}{\partial z}=0$)

$$⋮$$

Only when we look at $\frac{d}{dt}{c}^{(n-1)}=c^{(n)}(x_1,x_2,x_3,\dots ,x_n,{\dot{x}}_n)$, we realize that it is a function of ${\dot{x}}_n$, which again is a function of $z$. We can therefore now find $z$, and $\frac{\partial c^{(n)}}{\partial z}\ne 0$

Looking at $\frac{d}{dt}{c}^{(n)}=c^{(n+1)}(x,z,\dot{z})$, we realize that we get an ODE.

Note

The differential index of this example is therefore $n$.

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